Re: Normal Forms and BCNF

> Hi,> I am a little confused with Normal Forms and BCNF.> I have some relations and I am trying to determine it's normal form. I> also need to decompose it to BCNF forms...Any help would be> appreciated.>> Here are some of the relations:> 1. R(A, B, C, D) where A -> B> B -> C, D

Since A->B and B->CD we have:

A->CD and A->BCD (hint: Armstrong's rules)

Conclusion: A is a candidate key (A is also the only candidate key).
Relation R IS NOT in BCNF because of the functional dependence B->CD (B is
not a candidate key). Relation R IS NOT in 3NF because of the functional
dependence B->CD (neither B is a candidate key nor CD is a subset of a
candidate key). Relation IS in 2NF because no non-key attribute depends on a
part of a candidate key.

I hope you are familiar with the algorithm for normalizing into BCNF. Since
functional dependence B->CD violates the condition for relation to be in
BCNF we decompose relation R into:

R1(B, C, D) KK={B}
R'=(A, B)=R2 KK={A}

R1 is in BCNF, R2 too. So, the final answer is:

R1(B, C, D) KK={B}
R2(A, B) KK={A}

Note that we can reconstruct original relation R by joining relations R1 and
R2 on R1.B = R2.B